I have tried searching the literature for a result like the following, but have not found anything.
For a positive integer $m$, is it known that $$\{ \sin (k \pi / m): 1 \leq k \leq m/2, (k,m)=1 \}$$ is linearly independent over the rationals?
References or a proof would be greatly appreciated.
Note: Fedor and Vladimir have already answered the question, but this is a partial answer in the other direction, under a stronger hypothesis. (This answer, which I had earlier deleted, has been edited in response to some helpful comments.)
If $m$ is odd and square-free, then the claim of the OP holds.
Let $S$ be the set $\{k: 1 \leq k \leq m/2, (k, m) = 1\}$. If $\sum_{k \in S} a_k \sin(k\pi/m) = 0$ for some rationals $a_k$, then
$$\sum_{k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) = 0.$$
Here $N = 2m$ is square-free, and in that case the primitive $N$-th roots of unity are linearly independent over $\mathbb{Q}$ (see this mathstackexchange discussion: https://math.stackexchange.com/questions/87290/basis-of-primitive-nth-roots-in-a-cyclotomic-extension).
For $N = 2m$, any odd $k \in S$ is prime to $N$ and hence $e^{k \pi i/m}, e^{-k\pi i/m}$ are primitive $N$-th roots of unity. If $k \in S$ is even, then $m + k$ is odd and prime to $m$ and thus to $N$, so $e^{(m + k)\pi i/m} = -e^{k\pi i/m}$ is also primitive $N$-th root of unity, as is its conjugate $e^{(m - k)\pi i/m}$; notice the $m-k$ lie in $\{j: m/2 \leq j \leq m, \gcd(j,m) = 1\}$ which is disjoint from $S$. Then
$$\begin{array}{lll} \sum_{k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) & = & \sum_{\text{odd}\; k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) + \sum_{\text{even}\; k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) \\ & = & \sum_{\text{odd}\; k \in S} a_k(e^{k\pi i/m} - e^{-k\pi i/m}) + \sum_{\text{even}\; k \in S} a_k(e^{(m-k)\pi i/m} - e^{-(m-k)\pi i/m}) \end{array}$$
where all the primitive roots of unity appearing in the last expression are manifestly distinct. By linear independence of the primitive roots, if that linear combination is zero, then $a_k = 0$ for all $k$, as required.
$$\sin \frac{5\pi}{36}+\sin \frac{7\pi}{36}-\sin \frac{17\pi}{36}=0.$$ This may be otained by multiplying $2\sin\frac{\pi}6-1=0$ by $\cos \frac{\pi}{36}$.
We have $$\sin\frac{\pi}{9}+\sin\frac{2\pi}9-\sin\frac{4\pi}9=\sin\frac{2\pi}{18}+\sin\frac{4\pi}{18}-\sin\frac{8\pi}{18}=\sin\frac{2\pi}{18}-\sin\frac{8\pi}{18}+\sin\frac{14\pi}{18},$$ and the latter, denoting $\xi_{18}=\exp\frac{2\pi i}{18}$, is the imaginary part of $$\xi_{18}-\xi_{18}^4+\xi_{18}^7=\xi_{18}(1-\xi_{18}^3+\xi_{18}^6)=0.$$ Thus, your conjecture is wrong.
